Problem 73 - A Complete Resource Book in Mathematics for JEE Main

Question

The differential equation corresponding to \(y=\sum_{i=1}^{3} c_{i} e^{m, x}\), where \(c_{i}\) 's are arbitrary constants and \(m_{1}, m_{2}, m_{3}\) are roots of the equation \(m^{3}-7 m+6=0\), is (A) \(y_{3}-7 y_{1}+6 y=0\) (B) \(y_{3}+7 y_{1}+6 y=0\) (C) \(y_{3}-7 y_{1}-6 y=0\) (D) None of these

Step-by-Step Solution

Answer

The correct option is (A) \(y_{3} - 7 y_{1} + 6 y = 0\).

Step 1: Identify the Roots of the Given Equation

First, find the roots of the polynomial equation \(m^3 - 7m + 6 = 0\). We check possible rational roots using the Rational Root Theorem and find that \(m = 1\) is a root. We can use synthetic division or polynomial division to factor \(m^3 - 7m + 6\) as \((m - 1)(m^2 + m - 6)\). The quadratic \(m^2 + m - 6\) can be factored further as \((m - 2)(m + 3)\), giving roots \(m = 2\) and \(m = -3\). Thus, the roots are \(m_1 = 1\), \(m_2 = 2\), and \(m_3 = -3\).

Step 2: Form the General Solution

Given the roots \(m_1 = 1\), \(m_2 = 2\), and \(m_3 = -3\), the complementary function (general solution) of the corresponding differential equation is \(y = c_1 e^{x} + c_2 e^{2x} + c_3 e^{-3x}\).

Step 3: Determine the Differential Equation

The general solution \(y = c_1 e^{x} + c_2 e^{2x} + c_3 e^{-3x}\) corresponds to a homogeneous linear differential equation with constant coefficients. The characteristic equation is \((m - 1)(m - 2)(m + 3) = 0\), which expands to \(m^3 - 7m + 6 = 0\). The differential equation formed from this is the third-order derivative equation \(y^{(3)} - 7y' + 6y = 0\).

Step 4: Match with Given Options

Compare the derived differential equation \(y^{(3)} - 7y' + 6y = 0\) with the given options. In these options, \(y_3\) denotes the third derivative, \(y_1\) denotes the first derivative, and \(y\) is the function itself. Match it with option (A) which is \(y_{3} - 7y_{1} + 6y = 0\).

Key Concepts

Polynomial Roots

When dealing with polynomial equations, finding the roots is a crucial step. These roots are the values of the variable that make the polynomial equal to zero. In our exercise, we have the polynomial equation \(m^3 - 7m + 6 = 0\). To find the roots, we can apply the Rational Root Theorem, which helps us test possible rational roots derived from the coefficients of the polynomial.

• First, test \(m = 1\) and check if it makes the polynomial zero. Since it works, \(m = 1\) is a root.
• Next, we use synthetic division or polynomial division to simplify the polynomial using this root.
• The result is \((m - 1)(m^2 + m - 6)\).
• We further factor \(m^2 + m - 6\) to get the roots \(m = 2\) and \(m = -3\).

Thus, the roots of the polynomial are \(m_1 = 1\), \(m_2 = 2\), and \(m_3 = -3\). These roots are foundational for deriving the general solution of the differential equation.

Characteristic Equation

The characteristic equation is a polynomial equation derived from a differential equation that helps to find solutions of the form \(y = e^{mx}\). These solutions are specifically suitable for linear homogeneous differential equations with constant coefficients.
For the differential equation, we establish the characteristic equation by assuming a solution that involves exponential functions. This equation is derived from substituting the assumed solutions back into the differential equation context.
In our problem, the characteristic equation is formed from the polynomial equation \((m-1)(m-2)(m+3) = 0\), which expands to \(m^3 - 7m + 6 = 0\). The roots of this characteristic equation, \(m_1 = 1\), \(m_2 = 2\), and \(m_3 = -3\), indicate the shape and form of the general solution of the differential equation. This equation plays a pivotal role in determining the complete solution structure.

Linear Homogeneous Equations

Linear homogeneous differential equations are a type of differential equation where all terms are a multiple of the function or its derivatives, and the equation equals zero. This means that in any given equation of this type, there are no standalone (non-zero) constants present.
The solutions to these equations often take the form of exponential functions, leading to simple combinations of such functions forming the general solution.
In our given problem, the differential equation is \(y^{(3)} - 7y' + 6y = 0\). This reflects the general form of a linear homogeneous equation where:

• \(y^{(3)}\) represents the third derivative of the function,
• \(y'\) is the first derivative, and
• \(y\) is the function itself.

These terms all contain the dependent variable or its derivatives, adhering to the principle of homogeneity by equating everything to zero. Through solving such equations, we derive forms like \(y = c_1 e^{x} + c_2 e^{2x} + c_3 e^{-3x}\), which serve as the general solution based on the polynomial roots obtained from the characteristic equation. Understanding this concept helps in constructing solutions that satisfy the differential equation throughout its domain.